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UA OPTI512R 傅立叶光学导论15 2-D Fourier变换与Hankel变换

發(fā)布時(shí)間:2025/4/14 编程问答 36 豆豆
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UA OPTI512R 傅立葉光學(xué)導(dǎo)論15 2-D Fourier變換與Hankel變換

    • 2-D Fourier變換的定義
    • 2-D Fourier變換的性質(zhì)
    • 極坐標(biāo)系中的2-D Fourier變換
    • Hankel變換

2-D Fourier變換的定義

定義 2-D Fourier變換
F(ξ,η)=F[f(x,y)]=∫?∞+∞f(x,y)e?j2π(ξx+ηy)dxdyF(\xi,\eta)=\mathcal{F}[f(x,y)]=\int_{-\infty}^{+\infty} f(x,y)e^{-j 2 \pi (\xi x+\eta y)}dxdyF(ξ,η)=F[f(x,y)]=?+?f(x,y)e?j2π(ξx+ηy)dxdy

2-D Fourier逆變換
f(x,y)=F?1[F(ξ,η)]=∫?∞+∞F(ξ,η)ej2π(ξx+ηy)dξdηf(x,y)=\mathcal{F}^{-1}[F(\xi,\eta)]=\int_{-\infty}^{+\infty} F(\xi , \eta)e^{j 2 \pi (\xi x+\eta y)}d\xi d \etaf(x,y)=F?1[F(ξ,η)]=?+?F(ξ,η)ej2π(ξx+ηy)dξdη

這個(gè)定義要推廣到NNN維也是很容易的,
F(ξ?)=F[f(x?)]=∫?∞+∞f(x?)e?j2πξ??x?dx?f(x?)=F?1[F(ξ?)]=∫?∞+∞F(ξ?)ej2πξ??x?dξ?F(\vec \xi)=\mathcal{F}[f(\vec x)]=\int_{-\infty}^{+\infty} f(\vec x)e^{-j 2 \pi \vec \xi \cdot \vec x}d\vec x \\ f(\vec x)=\mathcal{F}^{-1}[F(\vec \xi)]=\int_{-\infty}^{+\infty} F(\vec \xi)e^{j 2 \pi \vec \xi \cdot \vec x}d\vec \xiF(ξ?)=F[f(x)]=?+?f(x)e?j2πξ??xdxf(x)=F?1[F(ξ?)]=?+?F(ξ?)ej2πξ??xdξ?

例1 可分函數(shù)(separable function):
f(x,y)=g(x)h(y)f(x,y)=g(x)h(y)f(x,y)=g(x)h(y)

它的Fourier變換為
F(ξ,η)=∫?∞+∞f(x,y)e?j2π(ξx+ηy)dxdy=∫?∞+∞g(x)h(y)e?j2π(ξx+ηy)dxdy=(∫?∞+∞g(x)e?j2πξxdx)(∫?∞+∞h(y)e?j2πηydy)=F[g(x)]F[h(y)]\begin{aligned} F(\xi,\eta) & =\int_{-\infty}^{+\infty} f(x,y)e^{-j 2 \pi (\xi x+\eta y)}dxdy \\ & = \int_{-\infty}^{+\infty} g(x)h(y)e^{-j 2 \pi (\xi x+\eta y)}dxdy \\ & = \left( \int_{-\infty}^{+\infty} g(x)e^{-j 2 \pi \xi x}dx \right) \left( \int_{-\infty}^{+\infty} h(y)e^{-j 2 \pi \eta y}dy \right) \\ & = \mathcal{F}[g(x)]\mathcal{F}[h(y)]\end{aligned}F(ξ,η)?=?+?f(x,y)e?j2π(ξx+ηy)dxdy=?+?g(x)h(y)e?j2π(ξx+ηy)dxdy=(?+?g(x)e?j2πξxdx)(?+?h(y)e?j2πηydy)=F[g(x)]F[h(y)]?

對(duì)于一般函數(shù),
F(ξ,η)=∫?∞+∞f(x,y)e?j2π(ξx+ηy)dxdy=∫?∞+∞f(x,y)e?j2π(ξx+ηy)dxdy=∫?∞+∞(∫?∞+∞f(x,y)e?j2πξxdx)e?j2πηydy=Fy[Fx[f(x,y)]]\begin{aligned} F(\xi,\eta) & =\int_{-\infty}^{+\infty} f(x,y)e^{-j 2 \pi (\xi x+\eta y)}dxdy \\ & = \int_{-\infty}^{+\infty} f(x,y)e^{-j 2 \pi (\xi x+\eta y)}dxdy \\ & = \int_{-\infty}^{+\infty}\left( \int_{-\infty}^{+\infty} f(x,y)e^{-j 2 \pi \xi x}dx \right) e^{-j 2 \pi \eta y}dy \\ & = \mathcal{F}_y[\mathcal{F}_x[f(x,y)]]\end{aligned}F(ξ,η)?=?+?f(x,y)e?j2π(ξx+ηy)dxdy=?+?f(x,y)e?j2π(ξx+ηy)dxdy=?+?(?+?f(x,y)e?j2πξxdx)e?j2πηydy=Fy?[Fx?[f(x,y)]]?

也就是說(shuō)多元函數(shù)的Fourier變換是可以序貫地對(duì)每個(gè)變量做Fourier變換的。

2-D Fourier變換的性質(zhì)

性質(zhì)原函數(shù)Fourier變換
Scalingf(ax,by)f(ax,by)f(ax,by)1abF(ξ/a,η/b),ab>0;?1abF(ξ/a,η/b),ab<0\frac{1}{ab}F(\xi/a,\eta/b),ab>0; -\frac{1}{ab}F(\xi/a,\eta/b),ab<0ab1?F(ξ/a,η/b),ab>0;?ab1?F(ξ/a,η/b),ab<0
Shiftingf(x?x0,y?y0)f(x-x_0,y-y_0)f(x?x0?,y?y0?)e?j2π(ξx0+ηy0)F(ξ,η)e^{-j2 \pi (\xi x_0+\eta y_0)}F(\xi ,\eta)e?j2π(ξx0?+ηy0?)F(ξ,η)
卷積定理f(x,y)?h(x,y)f(x,y)*h(x,y)f(x,y)?h(x,y)F(ξ,η)H(ξ,η)F(\xi,\eta)H(\xi,\eta)F(ξ,η)H(ξ,η)
卷積定理(頻域)F(ξ,η)?H(ξ,η)F(\xi,\eta)*H(\xi,\eta)F(ξ,η)?H(ξ,η)f(x,y)h(x,y)f(x,y)h(x,y)f(x,y)h(x,y)

其中2-D卷積的定義是
f(x,y)?h(x,y)=∫?∞+∞f(α,β)h(x?α,y?β)dαdβf(x,y)*h(x,y)=\int_{-\infty}^{+\infty} f(\alpha,\beta)h(x-\alpha,y-\beta)d \alpha d \betaf(x,y)?h(x,y)=?+?f(α,β)h(x?α,y?β)dαdβ

極坐標(biāo)系中的2-D Fourier變換

從定義出發(fā),用二重積分的換元公式得到極坐標(biāo)系中的2-D Fourier變換,
F(ξ,η)=F[f(x,y)]=∫?∞+∞f(x,y)e?j2π(ξx+ηy)dxdyF(\xi,\eta)=\mathcal{F}[f(x,y)]=\int_{-\infty}^{+\infty} f(x,y)e^{-j 2 \pi (\xi x+\eta y)}dxdyF(ξ,η)=F[f(x,y)]=?+?f(x,y)e?j2π(ξx+ηy)dxdy

(x,y)(x,y)(x,y)變換到極坐標(biāo)(r,θ)(r,\theta)(r,θ)中,將(ξ,η)(\xi,\eta)(ξ,η)變換到極坐標(biāo)(ρ,?)(\rho,\phi)(ρ,?)中,即
{x=rcos?θy=rsin?θ,{ξ=ρcos??η=ρsin??\begin{cases} x = r \cos \theta \\ y = r \sin \theta \end{cases},\begin{cases} \xi = \rho \cos \phi \\ \eta = \rho \sin \phi \end{cases}{x=rcosθy=rsinθ?,{ξ=ρcos?η=ρsin??

所以
xξ+yη=rρ(cos?θcos??+sin?θsin??)=rρcos?(θ??)x \xi + y \eta = r \rho(\cos \theta \cos \phi + \sin \theta \sin \phi) = r \rho \cos(\theta -\phi)xξ+yη=rρ(cosθcos?+sinθsin?)=rρcos(θ??)

Jacobian為
∣?x?r?x?θ?y?r?y?θ∣=∣cos?θrsin?θsin?θ?rcos?θ∣=?r\left| \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{matrix} \right|=\left| \begin{matrix} \cos \theta & r \sin \theta \\ \sin \theta& -r \cos \theta\end{matrix} \right|=-r??r?x??r?y???θ?x??θ?y???=?cosθsinθ?rsinθ?rcosθ??=?r

因此,對(duì)f(x,y)=f(x(r,θ),y(r,θ))f(x,y)=f(x(r,\theta),y(r,\theta))f(x,y)=f(x(r,θ),y(r,θ))根據(jù)積分換元公式
F(ρ,?)=∫?∞+∞f(x,y)e?j2π(ξx+ηy)dxdy=∫02π∫0+∞r(nóng)f(r,θ)e?j2πrρcos?(θ??)drdθ\begin{aligned} F(\rho,\phi)&=\int_{-\infty}^{+\infty} f(x,y)e^{-j 2 \pi (\xi x+\eta y)}dxdy \\ & =\int_0^{2\pi} \int_{0}^{+\infty} rf(r,\theta)e^{-j 2 \pi r \rho \cos(\theta - \phi)}dr d \theta \end{aligned}F(ρ,?)?=?+?f(x,y)e?j2π(ξx+ηy)dxdy=02π?0+?rf(r,θ)e?j2πrρcos(θ??)drdθ?

這就是極坐標(biāo)中的函數(shù)f(r,θ)f(r,\theta)f(r,θ)的Fourier變換公式。

Hankel變換

對(duì)于任意極坐標(biāo)系中的函數(shù)f(r,θ)f(r,\theta)f(r,θ),它可以展開(kāi)為
f(r,θ)=∑k=?∞+∞fk(r)ejkθf(wàn)(r,\theta)=\sum_{k=-\infty}^{+\infty} f_k(r)e^{jk\theta}f(r,θ)=k=?+?fk?(r)ejkθ

它的Fourier變換為
F(ρ,?)=∫02π∫0+∞r(nóng)f(r,θ)e?j2πrρcos?(θ??)drdθ=2π∑k=?∞+∞(?j)kejk?Hk[fk(r)]\begin{aligned} F(\rho,\phi)&=\int_0^{2\pi} \int_{0}^{+\infty} rf(r,\theta)e^{-j 2 \pi r \rho \cos(\theta - \phi)}dr d \theta \\ & =2\pi \sum_{k=-\infty}^{+\infty} (-j)^k e^{jk\phi} \mathcal{H}_k[f_k(r)]\end{aligned}F(ρ,?)?=02π?0+?rf(r,θ)e?j2πrρcos(θ??)drdθ=2πk=?+?(?j)kejk?Hk?[fk?(r)]?

假設(shè)f(r,θ)f(r,\theta)f(r,θ)是極坐標(biāo)系中的可分函數(shù),即
f(r,θ)=fR(r)fΘ(θ)f(r,\theta)=f_R(r)f_{\Theta}(\theta)f(r,θ)=fR?(r)fΘ?(θ)

則它的展開(kāi)式中fk=fRf_k=f_Rfk?=fR?,Fourier變換為
F(ρ,?)=∑k=?∞+∞ck(?j)kejk?Hk[fR(r)]\begin{aligned} F(\rho,\phi) & = \sum_{k=-\infty}^{+\infty} c_k (-j)^k e^{jk\phi} \mathcal{H}_k[f_R(r)]\end{aligned}F(ρ,?)?=k=?+?ck?(?j)kejk?Hk?[fR?(r)]?

這里ckc_kck?表示Fourier級(jí)數(shù)展開(kāi)的系數(shù),
ck=12π∫02πfΘ(θ)e?jkθdθc_k=\frac{1}{2\pi}\int_0^{2\pi}f_{\Theta}(\theta)e^{-jk\theta}d \thetack?=2π1?02π?fΘ?(θ)e?jkθdθ

Hk\mathcal{H}_kHk?表示kkk階Hankel變換,
Hk[fR(r)]=2π∫0+∞r(nóng)fR(r)Jk(2πrρ)dr\mathcal{H}_k[f_R(r)]=2 \pi \int_0^{+\infty} r f_R(r)J_k(2 \pi r \rho) drHk?[fR?(r)]=2π0+?rfR?(r)Jk?(2πrρ)dr

其中JkJ_kJk?表示kkk階第一類Bessel函數(shù)。

Radially Symmetric Function
假設(shè)極坐標(biāo)系中的可分函數(shù)滿足fΘ=1f_{\Theta}=1fΘ?=1,就稱其為Radially Symmetric Function,它的Fourier變換為
F(ρ,?)=∫02π∫0+∞r(nóng)f(r,θ)e?j2πrρcos?(θ??)drdθ=∫02π∫0+∞r(nóng)fR(r)e?j2πrρcos?(θ??)drdθ\begin{aligned} F(\rho,\phi)&=\int_0^{2\pi} \int_{0}^{+\infty} rf(r,\theta)e^{-j 2 \pi r \rho \cos(\theta - \phi)}dr d \theta \\ & =\int_0^{2\pi} \int_{0}^{+\infty} rf_R(r)e^{-j 2 \pi r \rho \cos(\theta - \phi)}dr d \theta\end{aligned}F(ρ,?)?=02π?0+?rf(r,θ)e?j2πrρcos(θ??)drdθ=02π?0+?rfR?(r)e?j2πrρcos(θ??)drdθ?

因?yàn)?br /> J0(2πrρ)=12π∫02πe?j2πrρcos?(θ??)dθJ_0(2 \pi r\rho )=\frac{1}{2\pi}\int_0^{2\pi} e^{-j 2 \pi r \rho \cos(\theta - \phi)} d \thetaJ0?(2πrρ)=2π1?02π?e?j2πrρcos(θ??)dθ

用Fubini定理,
F(ρ,?)=F(ρ)=2π∫0+∞r(nóng)fR(r)J0(2πrρ)drF(\rho,\phi)=F(\rho)=2 \pi \int_0^{+\infty}r f_R(r) J_0(2 \pi r \rho)d rF(ρ,?)=F(ρ)=2π0+?rfR?(r)J0?(2πrρ)dr

這個(gè)變換為0階Hankel變換,也被稱為Fourier-Bessel變換,它的逆變換為
fR(r)=2π∫0+∞ρF(ρ)J0(2πrρ)dρf_R(r)=2 \pi \int_0^{+\infty}\rho F(\rho)J_0(2 \pi r \rho)d \rhofR?(r)=2π0+?ρF(ρ)J0?(2πrρ)dρ

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